Educational Codeforces Round 159 [Rated for Div. 2]
By awoo, history, 11 months ago, translation, In English

Hello Codeforces!

On Dec/03/2023 17:35 (Moscow time) Educational Codeforces Round 159 (Rated for Div. 2) will start.

Series of Educational Rounds continue being held as Harbour.Space University initiative! You can read the details about the cooperation between Harbour.Space University and Codeforces in the blog post.

This round will be rated for the participants with rating lower than 2100. It will be held on extended ICPC rules. The penalty for each incorrect submission until the submission with a full solution is 10 minutes. After the end of the contest, you will have 12 hours to hack any solution you want. You will have access to copy any solution and test it locally.

You will be given 6 or 7 problems and 2 hours to solve them.

The problems were invented and prepared by Adilbek adedalic Dalabaev, Ivan BledDest Androsov, Maksim Neon Mescheryakov, Roman Roms Glazov and me. Also, huge thanks to Mike MikeMirzayanov Mirzayanov for great systems Polygon and Codeforces.

Good luck to all the participants!

UPD: Editorial is out

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11 months ago, # |
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11 months ago, # |
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Hope to get back to expert after this round! GL everyone :)

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Hope that it'll be an amazing contest!

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11 months ago, # |
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What's the rating distribution for problems

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11 months ago, # |
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Good luck everyone and have fun ^_^

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11 months ago, # |
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Educational Rounds are Mathforces af. Gonna skip this one

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11 months ago, # |
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My semester final exam in 4th December and 6th December and so on... But, I surely participate this Round also the upcoming div 3.

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11 months ago, # |
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11 months ago, # |
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Inshaallah, In this round I will be Green.

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11 months ago, # |
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hope it will be fun

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11 months ago, # |
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Educational rounds always offer so much to learn... Hope to solve D today in contest..

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11 months ago, # |
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Hoping to become a pupil after this round

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11 months ago, # |
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11 months ago, # |
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11 months ago, # |
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After 127 contests and I'm still a NEWBIE gonna set a world record :(

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11 months ago, # |
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D >> C

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    11 months ago, # ^ |
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    But E < D imo, so always be sure to read at least one more problem than the one you're stuck on.

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11 months ago, # |
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Speedforces

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11 months ago, # |
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Can anyone show me the formula to solve the problem B please? For me problem C is easier :D

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    11 months ago, # ^ |
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    you can do binary search instead of coming up a formula directly :D

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      11 months ago, # ^ |
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      I can't see the idea to do binary search. Can you help me?

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        11 months ago, # ^ |
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        well you got to do is find the greedy function check(x) which checks if x is valid solution and then do classic binary search whicch find the smallest x such that check(x) is true.i am not so sure because i came up with a formula during the contest

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        11 months ago, # ^ |
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        you can binary search the minimum amount of days that Monocarp cannot rest, let's say $$$mid$$$ days, then Monocarp can earn $$$mid*l+\min(2*mid,c)*t$$$ points, where $$$c=\lfloor (n+6)/7 \rfloor$$$ is the number of tasks unlocked within $$$n$$$ days.

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          11 months ago, # ^ |
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          I think it should be c = [(mid+6)/7] (correct me if I'm wrong)

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            11 months ago, # ^ |
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            It has to be with $$$n$$$, because maximum number of tasks that you can do is always the same and equals to $$$⌊(n+6)/7⌋$$$.

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    11 months ago, # ^ |
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    For me B was trivial not C :((

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    11 months ago, # ^ |
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    there are 3 thing you can do in a day

    1. attending a lesson
    2. attending a lesson + one task
    3. attending a lesson + two task

    we have ((day -1) / 7 + 1) tasks unlocked.

    so we will do 3. for tasks/2 days

    then do 2. if tasks is odd

    then 1. for remaining point

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      11 months ago, # ^ |
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      Thank you very much

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      11 months ago, # ^ |
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      that's exactly what i did start with the day type 3 and then to 2 and then to 1

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    11 months ago, # ^ |
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    Binary search is probably an easier method, but I found a formula, so here it is:

    Note that I'm pretty sure this constant-time solution is slower than binary search lmao.

    235560283

    Explanation
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      11 months ago, # ^ |
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      How can you justify the claim you made in the first para, I could attend more lectures and do. no practical I can get potential 14*l every 2 weeks, instead of l+2t, we don't know any relation between l and t so how can you say that that is the optimal, maybe the testcases are such that t is large enough to make it optimal

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        11 months ago, # ^ |
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        You have to minimize the number of days but if you are completing all the lectures then you are not minimizing. It would be better to consider first all possible l+2t and then the residue points should be used against the number of extra days where we will only attend lectures.

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        11 months ago, # ^ |
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        By 'most efficient' I meant optimizing for time spent studying, not time passing. In other words, there is no way to earn more than l + 2t points in a single day, and these days cannot be done more often than once every 2 weeks.

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    11 months ago, # ^ |
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    this is my submission, i used math not binary search.

    https://codeforces.net/contest/1902/submission/235665532

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11 months ago, # |
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Problem D is difficult to implement.

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11 months ago, # |
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Problem F is first known as "[SCOI2016] 幸运数字", a problem from 7 years ago

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11 months ago, # |
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For F, why using xor basis gives me WA on test 54?

https://codeforces.net/contest/1902/submission/235578100

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    11 months ago, # ^ |
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    Your code gives WA on 

    2
1 2
1 2
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1 1 1

    because in your dfs function, you say if you are the root node, don't initialize xor basis.

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11 months ago, # |
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I might have just clutched problem D

Great success?
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11 months ago, # |
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D >> E,F

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11 months ago, # |
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4 submissions on B :)

Also E < D.

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11 months ago, # |
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If anyone solved E using string hashing, drop the solution here

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11 months ago, # |
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:( +1 penalty in C because the constraint 'x is a positive integer' got updated after I opened the link. Also, no idea why directly solving the inequality in B gives WA but a binary search approach passes.

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    11 months ago, # ^ |
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    When solving the inequality, did you consider the case where the number of unlocked tasks is odd?

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      11 months ago, # ^ |
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      Doesn't floor() take care of parity issues?

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        11 months ago, # ^ |
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        Opposite actually, because you need one more day to complete the leftover task. I did calculations, not bisearch.

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        11 months ago, # ^ |
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        I guess you would have to take ceil not floor

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11 months ago, # |
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Cluthed D with 2 mins left !!

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11 months ago, # |
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me getting +3 on B but 0 on C didn't include a '()' if you see my submission :(

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11 months ago, # |
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Why E with such tight constraints on ML with trie, was it not intended? Saw many hashing solutions passing instead, made me sad ;(

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11 months ago, # |
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Failed D just because I forgot to check whether set.lower_bound(l) was equal to set.end(). Submitted the moment submissions opened and got AC. Good thing it was "out of competition".

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11 months ago, # |
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Looking into my time spent to solve each problem, the most difficult one from A to E was problem B.

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11 months ago, # |
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after this round i dont hate anything more than math

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First time solved F during contest and finished debugging E 1 sec after contest was over.

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11 months ago, # |
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I fucking missed $$$D$$$ just by a few seconds , bro!!!

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11 months ago, # |
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Whats the idea behind E?

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    11 months ago, # ^ |
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    I built a trie of all strings where each node further keeps track of the number of strings in the subtrie, as well as the total length of the strings in the subtrie while excluding the common prefix that the node represents.

    Then for each string, I read the characters in reverse-order and traverse the trie, with each step denoting a collapse step, while carefully accumulating the contributions to the desired sum of the other branches (concatenation cases) by utilizing the stored values.

    My submission: 235592382 I used a 2D array for the trie, which was probably a terrible idea (almost broke the memory limit), but the logic should be sound. I can elaborate on the details further if you want.

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11 months ago, # |
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I think problem D should give more time, I am sure I use a right O(nlogn) algorithm and implement it in python， however it's always show "tle". 2000ms is too short for D!

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    11 months ago, # ^ |
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    it is simple to solve D in < 500ms if you use cpp

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    11 months ago, # ^ |
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    pypy isn't always faster than python. you can sometimes switch from pypy to regular python and won't tle

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      11 months ago, # ^ |
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      Damn, you passed tests even without fast IO. I had to use cpp

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11 months ago, # |
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Please why is my submission for F WA on test 9

My thought is divide and conquer with XOR-basis with $$$q\log^2A$$$ complexity

code for F

thanks very much!!

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11 months ago, # |
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Am I the only fool who spent 1 hour and a half to figure out the solution for C while solved E in merely 15 minutes...

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    11 months ago, # ^ |
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    I'm the fool who couldn't solve E in 1 hour 40 minutes yet solved C in the first 11 minutes.

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11 months ago, # |
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In problem D, does the reversed commands range represent the same path, but mirrored around some line parallel to y = x?

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    11 months ago, # ^ |
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    Nope. Well, it might always be mirrored around some line, but that line definitely isn't necessarily parallel to y = x. For example, if you keep going up, then reversing does nothing, and your mirror line is a vertical line, and similarly, going right only yields a horizontal mirror line.

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      11 months ago, # ^ |
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      I think it should be mirrored around some dot, but I don't think that's somehow necessary for solution

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      11 months ago, # ^ |
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      Hmmm... Is it the line from the start of path to its end? If so, we can always just find it and mirror given point to check if it is within given range.

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        11 months ago, # ^ |
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        Nope, that's not it either. For example, if you move in a square wave, e.g., URDRURDRURDR..., the mirror line is a vertical line in the middle, which doesn't contain the start or the end.

        If there is some geometry-based solution to solve this problem, it would likely be much harder than simply precomputing the forward and reverse paths separately (which is likely the intended approach), so I wouldn't recommend thinking too hard about these geometric properties.

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    11 months ago, # ^ |
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    Nope

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    11 months ago, # ^ |
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    I might be wrong with that thinking, but my base idea is that after flipping, the prefix and suffix of points remains unchanged. If so, we can somehow transform given point and check if it is within prefix, suffix, and given range.

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    11 months ago, # ^ |
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    You're on the right track. Let $$$p_i$$$ be the point we're on before executing the $$$i^{th}$$$ command. Now to reverse a segment $$$[l, r]$$$ of the path, shift the origin to $$$p_{r+1}$$$ and mirror all the points on this segment around $$$x=-y$$$, (the transform is $$$(x,y)\to(-x,-y)$$$). Then, shift the origin to $$$p_l$$$. If we treat the points as vectors, then the point before the $$$i^{th}$$$ command after mirroring (assuming $$$i$$$ is in the segment) is $$$p'_i = -(p_i - p_{r+1}) + p_l = p_l+p_{r+1}-p_i$$$

    Why does this work? If the last command in the segment was R, then the second last point would be to the left of the ending point, so mirroring about the ending point will make the second point of the new segment to the right, which is what we want. Similarly this can be shown for any position and command on the segment. Finally we need to start at the original segment's starting point, so we shift our new points to have $$$p_l$$$ as the origin.

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      11 months ago, # ^ |
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      But isn't it O(r-l) for every query?

      I thought we need to transform the point in query so that it would be possible to check it on straight path stored in set.

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        11 months ago, # ^ |
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        I made it $$$O(\log n)$$$ per query by using a map that counted the number of times I visited each point. For a query x y l r: - If I visit x y before the $$$l^{th}$$$ command or after the $$$r^{th}$$$ command, then I visit that point after reversing the subarray - Between the $$$l^{th}$$$ and the $$$r^{th}$$$ command, I check if I visit the point $$$(x_l + x_{r+1} - x, y_l + y_{r+1} - y)$$$.

        To do this I sorted all the queries by l, and then updated counts of x y outside each segment and the transformed point inside the segment by using something similar to a sweep line method.

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          11 months ago, # ^ |
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          Yes, you are right. Simple set of points without visit time is not enough.

          But two simple sets are enough :) One to track from beginning and one to track from the end. I'll check this idea.

          upd: seems like even two sets without proper command times are not enough

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11 months ago, # |
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Damn, it's so sad I couldn't solve C in time, it was really easy

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11 months ago, # |
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Will Hashing Not Work in E, I was getting WA on test-5 :(

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How to solve D without Mo's ?

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    11 months ago, # ^ |
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    Pre-calculate times when you visit each position when moving forwards and backwards, plus some prefix sums to calculate the x and y offsets, then you're done!

    Basically this, I have been heavily commenting my solution during the contest 235604170

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    11 months ago, # ^ |
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    After reversing, suppose :

    x = s1 + s2 + s3...sl-1 + sr + sr-1 + sr-2.. s(r-z+1)

    => x = prefix_horizontal[l-1] + prefix_horizontal[r]-prefix_horizontal[r-z]

    => prefix_horizontal[r-z] = prefix_horizontal[l-1] + prefix_horizontal[r] — x

    Similarly, prefix_vertical[r-z] = prefix_vertical[l-1] + prefix_vertical[r] — y

    hence there's an index l-1<=z<=r such that,

    prefix_horizontal[r-z] = prefix_horizontal[l-1] + prefix_horizontal[r] — x

    prefix_vertical[r-z] = prefix_vertical[l-1] + prefix_vertical[r] — y

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11 months ago, # |
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How do you check in D whether you reach the point during the reversed section? I just guessed.

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    11 months ago, # ^ |
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    If you are choosing it inside the reversed part a[l,...,r]. Then it is choosing a[1..l-1] and a prefix from reversed(a[l...r]) which is suffix from a[l...r]

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11 months ago, # |
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My screencast of solving A-F
I'm planning to discuss my solutions (~2 hrs from contest end time)

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11 months ago, # |
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Can anyone tell what is causing runtime error in my solution? 235592760

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    11 months ago, # ^ |
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    gcd should not be equal to 0 initlize it with A[1] — A[0]

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    11 months ago, # ^ |
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    If n = 1, then You are returning without reading the array.So, there will be misinput.

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11 months ago, # |
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abs() dont accept long long in c++ are there any workaround other than to manually calculate it? I got WA in Problem C because of this.

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11 months ago, # |
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Can anybody tell me where did i go wrong in the code? problem B : https://codeforces.net/contest/1902/submission/235598766

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guys i want your help in identifying problem in my code, https://codeforces.net/contest/1902/submission/235602795 this is my solution for C, i first calculate the gcd of consecutive elements and then select a(n+1) by binary search,

i am getting runtime error on test case 2 with exit code 3 particularly on

16

9 -10 -7 4 -8 10 -5 -1 -9 8 3 -2 2 5 0 -6

the thing is when i run my code on my computer i get the correct answer, i have tried almost everything but can't figure out the problem

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    11 months ago, # ^ |
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    read this comment

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      11 months ago, # ^ |
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      that abs() one? i don't think thats an issue, i tried with llabs it still gives runtime error, i think there's some fault in the judge

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    11 months ago, # ^ |
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    When $$$n == 1$$$, your function returns without actually reading the array value, so the program would mistakenly treat this array value as the array length for the next test case.

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11 months ago, # |
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Can someone tell me why 235600689 underflows/overflows but 235608975 does not?

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11 months ago, # |
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Few more contests like these and I may reach Specialist

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11 months ago, # |
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Can anyone tell why this code gives TLE?

https://codeforces.net/contest/1902/submission/235611299

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11 months ago, # |
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For problem C, I spent about 50 minutes just for the runtime error with n = 1, long long difs[n - 1], difs[0]. Lesson learned for me. So glad to finally solve it on time.

Thank you guys for the contest!

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11 months ago, # |
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For me B was really frustrating. C was very good though!

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11 months ago, # |
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For B. Getting Points, although a mathematical formula exists, you can use Binary Search for ease of implementation.

The total tasks created will always be $$$(n + 6)/7$$$.

Suppose you take $$$r$$$ rest days. Notice that it's always optimal to take the first $$$r$$$ days off (so that you never run out of tasks). Since there are $$$(n - r)$$$ working days, and each day you can do at most 2 tasks, the points gained is $$$min(n + 6/7, 2*(n - r))$$$. The points gained via lessons is of course $$$(n - r)*l$$$.

Submission

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11 months ago, # |
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Can anyone share a typical case where C can fail?

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11 months ago, # |
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This educational round was really educative, enjoyed it a lot :)

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Don't understand why to use binary search in B.

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    11 months ago, # ^ |
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    you don't have to use BS. But fixing the days makes the math easier to not mess up. It introduces new errors that may be caused by binary search, though. In my opinion, its about preference.

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11 months ago, # |
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.

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11 months ago, # |
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Hope to become Master tomorrow, first time solved all problems in a contest.

I solved today F in O(n log^3 n) and O(n log^2 n), but the solutions are running in almost the same execution time.

Upd: My O(n log^2 n) solution dropped from 3200 ms to 700 ms after I used vector::reserve in N vectors of size at most 20

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11 months ago, # |
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https://codeforces.net/contest/1902/submission/235605516 could someone tell me, why am i getting TLE here? It should work in O(nlogn) which should easily pass for 1e6, right?

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11 months ago, # |
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in problem D ,Can anybody provide me a idea on how to check within ranges from L to R

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11 months ago, # |
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Bad problemset. Couldn't solve one

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    11 months ago, # ^ |
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    really cannot believe you are blaming contest for your bad level you have never solved a single problem in any of your contests and you are still nagging

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11 months ago, # |
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Surprisingly, really good contest i would say. Thanks for the authors! I really liked problems C and B. Idea behind B was clear, but solving it in O(1) took a lot of time). C was tricky to find the idea, but if you figured out how to take last element you can easily write it. Hope one day will have time and passion to open D,E problems!

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11 months ago, # |
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I misread problem C and ended up giving up :< I hope I wont be kicked off Expert

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11 months ago, # |
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Solved E for the first time in a Div.2 contest. Used Trie Here is the code

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11 months ago, # |
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Can anybody give me some problems that are similar to B, please? I feel uneasy when I encounter such a problem.

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    11 months ago, # ^ |
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    just solve randomly rather than looking for problems similar to some particular problem

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11 months ago, # |
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What knowledge do D & E require?

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11 months ago, # |
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Could someone give an example on how my code failed ?! I can't imagine the test case .

[submission:Hacked C Solution ]

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11 months ago, # |
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I feel that isn't it impolite to write hash cross code to someone else

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11 months ago, # |
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My Nlog^2N solution for D

  1. Think in terms of displacements on the X and Y axis.
  2. For a query L,R — (x,y) either lies in the prefix, the suffix or the middle region.
  3. Traverse the string forward and note down the points/cell reached at ith index for each i.
  4. Build a segment tree(each node contains a vector of pair representing all displacements that are achieved via this path) over this forward cells array and then to check if (x,y) is reached either in prefix or suffix: query the tree from (1,L-1) and (R+1,N) for (x,y)
  5. To check in mid region: build a segment over reversed array. Now you need to query if at any point in reversed section you can achieve the required displacement.
  6. Required displacement: since you have already moved prefix(L-1) so required displacement is Target — prefix(L-1).
  7. Reversed section starts from N-R+1 and ends at N-L+1. The origin here is already displaced by reversed(N-R). So account for that.
  8. Query the reversed segment tree for required displacement.

Submission link: https://codeforces.net/contest/1902/submission/235635773

Segment tree library used: https://youtu.be/K-86mKNAsmU?si=bI_IBJVNU50Mjysf

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11 months ago, # |
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I solved one (A) question, will my rating increase?

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11 months ago, # |
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Can anyone give me some tips for the hell C? I'm totally confused by it

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    11 months ago, # ^ |
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    GCD is the key to success

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      11 months ago, # ^ |
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      Probably you are right.But I wonder who all the elements turn to in the end?

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        11 months ago, # ^ |
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        the current biggest or the newest element which is bigger than the current biggest

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11 months ago, # |
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hoping i will become pupil after this round

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11 months ago, # |
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i'm wrong

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11 months ago, # |
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Can anyone please explain how to do C? How to apply GCD there ?

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    11 months ago, # ^ |
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    recall that all the numbers are unique.. also, as x is +ve you cannot decrease any number by adding x to it.

    now, we need to reach a common number that can be achieved by adding x multiple(or zero) times to all the numbers. Also, assume that a is sorted initially, so we are willing to achieve:

    a[0]+C1.x, a[1]+C2*x, .... , a[n-1]+Cn*x which makes a[0]+C1.x = a[1]+C2*x = ..... = a[n-1]+Cn*x

    so what you can see here is that we need to maximize x, also, greedily we can see that best choice for that final number as of now would be a[n-1], so we see that for any i, a[n-1]-a[i] will be some Ci*x, so take gcd of all such terms and find x.

    then the probem remains to find optimal the An (try it yourself, if you cant I can help there).

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      11 months ago, # ^ |
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      Hi akshat,

      Initially how i thought was, the best option to make array equal is making the array elements equal to the max element present in array.

      So what i did is found the max in array, for ex : a[] = [1 -19 17 -3 -15], max = 17; as the difference among (max — a[i]) should be multiple of x, so i thought x will be in range x -->[1,16], For which ever x is possible i have collected minimum answer(traversed whole array to check x is possible or not, which ever x is possible collected max-a[i]/x)may be here i can use GCD to find x right ?.

      I think u are also saying to make array equal to max right ?

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        11 months ago, # ^ |
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        Yes the idea is same as mine. You can look at my code for any clarifications :)

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    11 months ago, # ^ |
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    You're adding same x to all elements, if difference between them is not multiple of x they'll never become equal.

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11 months ago, # |
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Why does my nlogn solution timeout for D? submission — 235579778

Update — So many other nlogn solution also getting TimeLimitExceeded , any explanation?

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11 months ago, # |
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What's wrong with my solution 235568240 for C ? Please help me where it is leading to TLE

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    11 months ago, # ^ |
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    unordered_map leads to O(n^2)

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    11 months ago, # ^ |
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    By the way, for your check perfect square function you can instead just use __builtin_popcount which returns the number of bits and check if that is equal to one.

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      11 months ago, # ^ |
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      huh ?? That's wrong ...

      1000 = 8 is not a perfect square 1001 = 9 is a perfect square

      We can just check if a number is a power of 2 from no of set bits = 1

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        11 months ago, # ^ |
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        oh I'm dumb. I thought it checked powers of 2 not perfect squares because I was just working on another problem and that was in my head.

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11 months ago, # |
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Why so late Editorial?

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11 months ago, # |
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F can be done in O(n*lognlogA + q*logAlogA) with centroid decomposition. while decomposing the tree, just solve all the queries whose path includes the current centroid. We can do this by rooting the current subtree in the current centroid and calculate the xor base for every path from a centroid to a node in the current subtree(O(nlognlogA) in total) and then we can answer a query in O(logAlogA) by just merging 2 xor bases.

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    11 months ago, # ^ |
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    hey, can you explain your decomposition part a bit?

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      11 months ago, # ^ |
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      Lets say that you are currently solving your problem for a set of nodes S and a set of queries Q. When solving for S and Q, if you find a centroid C in the tree that is consisted of nodes in S, then every query from Q will either contain C, or not contain C. Those who contain C we will solve by precalculating xor bases for every path C->X, where X is in S, and then to answer a query (X,Y,K) we can merge the 2 paths that we precalculated C->X and C->Y and get the answer to our query. Now we decompose the tree into new trees, every query in Q that didnt contain C will now belong completely to some of the new trees. So that means that now we have new subproblems (S1,Q1),(S2,Q2).... which we can solve recursively.

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11 months ago, # |
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my $$$O((n + q) \cdot log_2n \cdot (log_2a_i)^2)$$$ solution on problem F using binary lifting and xor-basis passed during the contest but got TLE on test case 90 on system tests. and an optimized one passed with time complexity $$$O((n \cdot log_2n + q) \cdot (log_2a_i)^2)$$$ after the contest. Is there any solution with better time complexity, like for example $$$O((n + q) \cdot log_2n \cdot log_2a_i)$$$?

UPD: stefanbalaz2 just posted one 2 minutes ago. ty and what a coincidence.

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    11 months ago, # ^ |
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    I have other solution too, using LCA binary lift. submission

    Idea: For each node from root you can mantain the height where a new number is added to the xor-basis and the number added, when you are building the base in a child, you need to try do add only numbers added in the parent, thus at most 20 numbers will be added and at most 20 operations will be needed to know if a number will be added or not. for queries, I need to get the height of lca and add all numbers from basis of the nodes u and v up to the height.

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      11 months ago, # ^ |
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      oh, I got it, thank you, and your solution is easier to code for me.

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11 months ago, # |
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Hi, can anyone tell me how to hack "String hashing" solutions, just give a link to an article.

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11 months ago, # |
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unrated？

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11 months ago, # |
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When will rating be updated?

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11 months ago, # |
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@awoo@BledDest Thank you very much for the contest. Could we please get the editorials and ratings change — many thanks!

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    11 months ago, # ^ |
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    Most contests are rated within minutes of system checks completion. This one had hacking session finished over 15 hours back and still not rated.

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11 months ago, # |
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why this contest do not affect to rating's ?

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11 months ago, # |
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Oh my stupid me... In problem C, I assumed we must choose A[n+1] > 0, then got confused why it wouldn't work :(

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11 months ago, # |
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When Editorial will be released?

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11 months ago, # |
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after how much time does the rating changes come for educational rounds?

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11 months ago, # |
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I'm sorry but when rating changes ????

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11 months ago, # |
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Can you guys tell me if my ratings could go up? 1902C - Insert and Equalize

Click here to see my blogs

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    11 months ago, # ^ |
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    you can add the CF-Predictor extension for that

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11 months ago, # |
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In problem E, I used string hash and chose modulus 131, got AC during the contest but got WA on test 24 after contest. I tried 135, WA again on test 98. Finally I chose modulus 29123 and got AC. Could anyone tell me how to avoid such situation ?

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    11 months ago, # ^ |
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    Always use double hashing

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      11 months ago, # ^ |
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      I used double hashing. Double hashing can be wrong if choosing unlucky modulus.

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        11 months ago, # ^ |
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        131, 137 is common module for hashing so tester use some test case where your code will get WA . So use those prime module that is not common. Obviously not use 135 as a module cz that not a prime.

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          11 months ago, # ^ |
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          Thanks. I think I refer to some code template which is not so good.

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    11 months ago, # ^ |
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    here is my wrong submission 235733205 and right submission 235733378

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    11 months ago, # ^ |
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    I understand now, 131 is not mod number, it is base number and should be prime.

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      11 months ago, # ^ |
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      Base not necessary is a prime, I use base number randomed in range $$$[0, MOD)$$$

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        11 months ago, # ^ |
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        That is a good choice.

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11 months ago, # |
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ratings please..

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11 months ago, # |
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When will the ratings change

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11 months ago, # |
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ratings?

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11 months ago, # |
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Go everyone "WHERE IS RATING!!!"

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11 months ago, # |
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Auto comment: topic has been updated by awoo (previous revision, new revision, compare).

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11 months ago, # |
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If you want to decrease my rating, do it quick :(

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11 months ago, # |
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Any ideas why is the contest showing up as unrated for me?

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11 months ago, # |
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Why unrated?

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11 months ago, # |
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Where are ratings???

MikeMirzayanov
awoo
BledDest

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    11 months ago, # ^ |
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    I think mike KIDNAP the little innocent ratings

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11 months ago, # |
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No rating update?

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11 months ago, # |
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For D. I was thinking of a modified version of binary search on the string (landing on mid-point, and then deciding whether the left half or the right half can land me on the target point) — by this, I can answer each query in $$$O(log n)$$$ time. (reversing of string is trivial)

Can this approach work?

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11 months ago, # |
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I have got a message that unfortunately my solution 235585909 for problem C coincides with another solution. I don't know how did it happen. I used IDEONE. Maybe that could be the reason.

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11 months ago, # |
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problem F https://www.luogu.com.cn/problem/solution/P3292

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11 months ago, # |
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Regarding the mail -> Your solution 235540146 for the problem 1902A significantly coincides with solutions agrahariprajjawal5/235540146, iit2022164/235540396. Such a coincidence is a clear rules violation.

This has happened because of the use of a common source published before the competition. It happened because of the same template I and the other fellow user use. The template is taken from some internet source which I don't exactly remember. Unfortunately, the code inside the solve function also matched which is a mere coincidence. It's not strange that such small piece of code get matched. So, this is a mere coincidence and please revoke the submissions of mine done during the contest. I agree on providing more clarification regarding this.

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11 months ago, # |
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guys after the rating recalculations this contest was marked unrated for me while I'm still a newbie and haven't gone to 2100 to make it unrated why?

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11 months ago, # |
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Can anyone help me in a OA? I'll pay

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